Contents 11 Techniques of Integration

168

11.1 Basic Substitution and Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 11.2 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 11.3 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176

CONTENTS

2

Chapter 11

Techniques of Integration

11.1

Basic Substitution and Formulas

Definition The indefinite integral of a function f , written Z

f (x)dx,

is the most general antiderivative of f ; that is, Z f (x)dx = F (x) + C

(C is any constant)

if and only if F 0 (x) = f (x). Basic Formulas:

(Memorize this list)

Z

un+1 u du = +C n+1

Z Z

Z

du = ln |u| + C, u

eu du = eu + C

Z

au du =

sin u du = − cos u + C

Z

cos u du = sin u + C

csc2 u du = − cot u + C

n

(n 6= −1)

if

au +C ln a

Z

sec u du = tan u + C

Z

Z

sec u tan u du = sec u + C

Z

csc u cot u du = − csc u + C

Z

du √ = sin−1 u + C 1 − u2

Z

du = tan−1 u + C 1 + u2

2

u 6= 0

169

Z

11.1 Basic Substitution and Formulas

du √ = sec−1 u + C, |u| u2 − 1

|u| > 1

Z

tan u du = − ln | cos u| + C

sec u du = ln | sec u + tan u| + C

[f (u) + g(u)du] =

Z

cot u du = ln | sin u| + C

Z

Z

csc u du = ln | csc u − cot u| + C

Z

Z

kf (u)du = k

Z

f (u)du

1.

Z

(x2 + a)2 dx

3.

Z

5.

Z

ln | sec u| + C

or

f (u)du +

Z

g(u)du

if k is a constant

2.

Z

(x − 3)2 dx

x 2 dx

4.

Z

aπ dx x

Z

[(x + 1)3 + ex ]dx

6.

Z

x2 ex + x dx x2

7.

Z

10 sec2 x dx

8.

Z

16(2x + 1)dx

9.

Z

(sin x + cos x)dx

10.

Z

√ ( x + sec x tan x)dx

11.

Z

12.

Z

3x 3 1 x + + 2 2 x

13.

Z 

14.

Z

15.

Z

√ x x dx

16.

Z 

17.

Z

x √ dx x

18.

Z

x5 + x3 + 2 dx (Hint: divide) 1 + x2

19.

Z

x3 − a3 dx x−a

20.

Z

(x3 + 27)dx x2 − 3x + 9

21.

Z

1 − x2 dx 1 − x4

22.

Z 

1 x+ √ 1 − x2

23.

Z

24.

Z 

x3 − 8 x2 + 2x + 4

25.

Z

26.

Z

5

2

(sec x + 1)dx 4 3 5x − 3 + 5 x x





dx

 x3 a + x2 + a2 + x + A 3 2

 −1  tan x + c

! √ x 1 − x2 + 4 √ dx 1 − x2 x3 + x2 − x − 1 dx x−1

2

2

√

!

dx

x dx

√ 1 x x− √ x

dθ 2θ



dx





dx

dx

Techniques of Integration

170

27.

Z p

29.

Z

31.

Z

33.

Z

(x + 1)dx x2 + 2x + 2

34.

Z

2dy 3y − 4

35.

Z

4dt 5t + 2

36.

Z

x2 dx x3 + 4

37.

Z

(2x − 5)dx x2 − 5x + 3

38.

Z

v dv 6v 2 − 1

39.

Z

x3 dx x−1

40.

Z

(y − 3)dy y 2 − 6y + 1

41.

Z

x+6 dx (x + 2)2

42.

Z

y dy (1 + y 2 )4

43.

Z

√

44.

Z

v+3 dv v−1

45.

Z

√ x x + 1dx

46.

Z

x3 dx (x4 + 16)2

47.

Z

dt √ 1+ t

48.

Z

(x2 + 1)2 dx x3

49.

Z

(1 − 2x)2 dx x

50.

Z

cos β dβ 2 + 3 sin β

51.

Z

csc2 x dx 1 + cot x

52.

Z

sin 2t dt 4 − 3 cos 2t

53.

Z

cos x dx sin x

54.

Z

sin2 θ cos θ dθ

55.

Z

√ sin θ cos θ dθ

56.

Z

√ sin x √ dx x

57.

Z

x sin x2 dx

58.

Z

sec2 2t dt 1 + tan 2t

59.

Z

cos x dx sin4 x

60.

Z

sin 4x cos 4x dx

61.

Z

62.

Z

sec5 x tan x dx

2px dx 1

(1 + x2 ) 2 x dx x dx 3

(x2 − 1) 2

dx √ x(1 + x)

sin t dt 3

(4 + cos t) 2

 Hint:

x+6 (x + 2) + 4 = 2 (x + 2) (x + 2)2



28.

Z

dθ sec θ

30.

Z

x2 (2x3 + 1)2 dx

32.

Z

x dx 3x2 + 1

171

11.1 Basic Substitution and Formulas

√

63.

Z

e x √ dx x

64.

Z

x3 e2x dx

65.

Z

ey dy ey − 1

66.

Z

e−x dx 1 + e−x

67.

Z

dx x e +1

68.

Z

ex + 1 dx ex − 1

69.

Z

dx −x e + ex

70.

Z

et sec et tan et dt

71.

Z

sin(ln x)dx x

72.

Z

esin θ cos θ dθ

73.

Z

eln x dx x

74.

Z

eln x dx

75.

Z

ln x dx x

76.

Z

ex (4 − ex ) 2 dx

77.

Z

dx x(1 + 2 ln x)

78.

Z

x2 ln3 (1 + x3 ) dx 1 + x3

79.

Z

(1 + 2 ln x)4 dx x

80.

Z

tan θ ln(cos θ)dθ

81.

Z

(ln y) 2 dy y

82.

Z

3

√

dx 6x − 4x2

Solution to #82:

Let I =

6x − 4x2 = −4 x2 − ∴I=

Z

Now I =

1 2

x+

9 16

√


dx 6x − 4x2

+

9 4

=

9 4

−4 x−

dx q

9 4

−4 x− 3 4




q

9 4

Let u = 2 x −

∴I=

3 2

Z

1 2

Z

sin−1


3 2 4

∴ du = 2dx

du

4x−3 3

− u2 +C

=

1 u sin−1 +C 2 3/2


3 2 4

2

3

Techniques of Integration

83.

Z

172

x dx x2 + x + 1

Solution for #83: Let H =

Z

x2 + x + 1 = (x2 + x + 14 ) +

3 4

x dx x2 + x + 1 = (x + 12 )2 + 34 ;

let u = x + 12 ; du = dx; Z

then H =

(u − 12 )du = u2 + 34

Z

u du 1 3 − 2 2 u +4

In the first integral let w = u2 + ∴

Z

u du 1 3 = 2 2 u +4

Also

Z

Z

3 4

Z u2 +

du  √ 2 3 2

and we have 2u du = dw

dw 1 = ln |w| + C1 w 2

du 2 −1  √ 2 = √ tan 3 3 u2 +



2u √ 3



+ C2

2

∴ H=

1 2

ln |x2 + x + 1| −

84.

Z

dx x2 + 3x + 5

85.

Z

3x2

86.

Z

√

dx 4x − x2

87.

Z

√

dx 9x − 4x2

88.

Z

dx 12x2 + 56x + 72

89.

Z

9x2

90.

Z

√

91.

Z

√

92.

Z

2x2

93.

Z

√

dx − 12x + 16

8dx − 12x + 20

√1 3

tan−1



2x+1 √ 3



+C

h√

3 6

1 2

2

3

√

− 2) + C

i

sin−1

8x−9 9

+C



tan−1

3x−2 4

+C



+C



tan−1

3 2 (x

dx 8x − 25x2 −x2

dx − 2x + 8

dx + 3x + 2

x dx 3 + 4x − 4x2

 −1 sin

x+1 3

173

11.1 Basic Substitution and Formulas

94.

Z

√

95.

Z

(3x − 1)dx 9x2 + 6x + 26

96.

Z

√

97.

Z

dx √ (3x + 1) 3x2 + 2x − 5

98.

Z

dx √ (x + 1) x2 + 2x − 8

99.

Z

sin 2 θ cos3 θ dθ

100.

Z

cos− 2 θ sin θ dθ

101.

Z

cos3 x sin3 x dx

102.

Z

sin θ cos θ dθ

103.

Z

cos3 3θ dθ

104.

Z

sin3 x dx cos2 x

105.

Z

cos3 x csc3 x dx

106.

Z

sin3

107.

Z

tan2 x sec4 x dx

108.

Z

tan3 θ sec4 θ dθ

109.

Z

tan3 θ sec 2 θ dθ

110.

Z

(tan2 θ + tan4 θ)dθ

111.

Z

cot2

112.

Z

csc4 αx dx

 √ − 4x − x2 + sin−1

x dx 4x − x2

x−2 2

+C



4 3x+1

+C

i

sin 2 θ + C

i

sin2 θ + C1 = − 12 cos2 θ + C2 = − 14 cos 2θ + C



x dx 6 + x − 2x2 h

h

1

2 3

1 √

4 3

sec−1

3

sin 2 θ −

7

2 7

3

1 2

x dx 2

1 6

sec6 θ −

1 4

sec4 θ + C



cot αx + C



3

x dx 3  1 − α cot αx −

1 3α

Techniques of Integration

113.

Z

cot x csc2 x dx

114.

Z

sin6 x dx

115.

Z

sin4 x cos2 x dx

116.

Z

tan2 x cos4 x dx

117.

Z

sin4 x cos4 x dx

118.

Z

cos 2x sin 3x dx

119.

Z

cos 5x cos 7x dx −

120.

Z

sin− 2 u cos3 u du

11.2

174

1 2

Z

1 2

sin x + C



tan−1 x − 12 x + C



sin 5x +

sin 5x sin 7x dx

5

Integration by Parts

121.

Z

x cos x dx

122.

Z

x sin 2x dx

123.

Z

x2 sin 2x dx

124.

Z

x2 dx sec x

125.

Z

tan−1 x dx

126.

Z

x tan−1 x dx

127.

Z

sin−1

128.

Z p

129.

Z

x a

dx, a > 0

1 − x2 dx

ln x dx

1

2 2x

tan−1 x +

1 2

175

11.2 Integration by Parts

130.

Z

x ln x dx

131.

Z

x2 ln x dx

132.

Z

x−1 ln x dx

133.

Z

sec3 θ dθ

134.

Z

cos

135.

Z

√

136.

Z

ln2 x dx

137.

Z

ex sin x dx

138.

Z

e−2x cos x dx

139.

Z

2x sin x dx

140.

Z

√ sin−1 x dx √ x

141.

Z

tan−1

142.

Z

x3 ex dx

143.

Z

x3 e2x dx

144.

Z

θ sec2 aθ dθ

145.

Z

x2 sin−1 x dx

146.

Z

sin3 x dx

147.

Z

cos 2x sin x dx

148.

Z

x2 e−x dx

√

y dy

y cos

√

√

y dy

  √ √ √ √ 2y sin y + 4 y cos y − 4 sin y + C

  √ √ (x + 1) tan−1 x − x + C

x dx

2

1

3 3x

√  sin−1 x + 19 (2 + x2 ) 1 − x2 + C

Techniques of Integration

149.

Z

sin θ ln(cos θ)dθ

150.

Z

xax dx

151.

Z

θ sin θ cos θ dθ

152.

Z

ln(ln x) dx x

153.

Z

√ ln x + a dx x+a

154.

Z

x3

11.3

176

[cos θ(1 − ln(cos θ)) + C]

1 4

ln2 (x + a) + C



p a2 − x2 dx

Improper Integrals

In the following problems determine whether or not each of the improper integrals is convergent, and compute its value if it is. Z a dx √ 155. 2 − x2 a 0 Solution for #155: Let A = lim

t→a

∴ A = lim sin−1 x→a

156.

2

dx x2

2

dx x2 − 4

Z

0

dx x+2

Z

1

Z 0

157.

Z 0

158.

−2

159.

0

160.

Z

1

0

161.

Z

162.

Z 0




=

0

dx √ x

t

√

dx − x2

a2

π 2

dx √ x

[2]

(use substitution u =

1

dx ,0